Answer to Question #124660 in Analytic Geometry for desmond

Question #124660
Verify that the point P(a cos θ, b sin θ) lies on the ellipse
x
2
a
2
+
y
2
b
2
= 1,
where a and b are the semi-major and semi-minor axes respectively of the ellipse . Find the
gradient of the tangent to the curve at P and show that the equation of the normal at P is
ax sin θ − by cos θ = (a
2 − b
2
) sin θ cos θ.
If P is not on the axes and if the normal at P passes through the point B(0, b), Show that
a
2 > 2b
2
. If further, the tangent at P meets the y-axis at Q, show that
|BQ| =
a
2
b
2
.
1
Expert's answer
2020-07-01T19:41:41-0400

If the point "P(acos\\theta,bsin\\theta)" satisfies the equation "\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1" , then it lies on the given ellipse .


Now "\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = \\frac{(acos\\theta)^2}{a^2}+\\frac{(bsin\\theta)^2}{b^2}"

"= \\frac{a^2cos^2\\theta}{a^2}+\\frac{b^2sin^2\\theta}{b^2}"

"= cos^2\\theta+sin^2\\theta"

"=1."

"\\therefore" point "P(acos\\theta,bsin\\theta)" lies on the given ellipse. (PROVED).


Now the equation of the tangent at the point "P(acos\\theta,bsin\\theta)" is given by

"\\frac{x*acos\\theta}{a^2}+\\frac{y*bsin\\theta}{b^2}=1\\\\or, \\space \\frac{xcos\\theta}{a}+\\frac{ysin\\theta}{b}=1\\\\or, \\space \\frac{bxcos\\theta+aysin\\theta}{ab}=1\\\\or, \\space bxcos\\theta+aysin\\theta=ab\\\\or, \\space aysin\\theta=-bxcos\\theta+ab\\\\or, \\space y=-\\frac{bxcos\\theta}{asin\\theta}+\\frac{ab}{asin\\theta}\\\\or, \\space y=-(\\frac{b}{a}tan\\theta)x +bcosec\\theta."


"\\therefore" Gradient of the tangent at the point "P(acos\\theta,bsin\\theta)" is "-(\\frac{b}{a}tan\\theta)" . Answer.


Equation of Normal: We know that normal is perpendicular to the tangent..


so if equation of tangent is "bxcos\\theta+aysin\\theta=ab" then equation of normal is of the form


"axsin\\theta-bycos\\theta=k, \\space where \\space k \\space is \\space constant."


Now as this normal is passing through "P(acos\\theta,bsin\\theta)" ...by putting this coordinate into the equation of normal we get the value of "k" .


"\\therefore a*acos\\theta*sin\\theta-b*bsin\\theta*cos\\theta=k\\\\or, \\space k=a^2sin\\theta cos\\theta-b^2sin\\theta cos\\theta\\\\or, \\space k= (a^2-b^2)sin\\theta cos\\theta"


"\\therefore" The equation of normal at the point "P(acos\\theta,bsin\\theta)" is given by

"axsin\\theta-bycos\\theta= (a^2-b^2)sin\\theta cos\\theta." PROVED.


Now if the equation of normal is passing through B(0,b) then we have

"a*0*sin\\theta-b*b*cos\\theta=(a^2-b^2)sin\\theta cos\\theta\\\\or, \\space -b^2cos\\theta=(a^2-b^2)sin\\theta cos\\theta.\\\\ or, \\space -b^2=(a^2-b^2)sin\\theta \\\\or, \\space sin\\theta = \\frac{-b^2}{a^2-b^2}"


Now as we know "-1\\leq sin\\theta,\\\\"

"\\therefore -1\\leq \\frac{-b^2}{a^2-b^2}\\\\or, 1\\geq \\frac{b^2}{a^2-b^2}\\\\ or, a^2-b^2 \\geq b^2\\\\or, a^2 \\geq2b^2" PROVED.


Now tangent at "P(acos\\theta,bsin\\theta)" which is "bxcos\\theta+aysin\\theta=ab" , meets the Y axis at Q.

putting x=0 ,we get

"0+aysin\\theta=ab\\\\or, \\space ysin\\theta=b\\\\or, \\space y=\\frac{ b}{sin\\theta}.\\\\"


"\\therefore" The coordinate of Q is "(0,\\frac{b}{sin\\theta})" .


"BQ= \\bigg(\\frac{b}{sin\\theta}-b\\bigg)\\\\BQ=\\bigg(\\frac{b}{\\frac{-b^2}{a^2-b^2}}-b\\bigg)"

"BQ= b(\\frac{b^2-a^2}{b^2}-1)\\\\BQ= b*(\\frac{b^2-a^2-b^2}{b^2})\\\\BQ=\\frac{-a^2}{b}"

"\\therefore |BQ|= \\frac{a^2}{b}" PROVED.


After solving question i found that BQ is "\\frac{a^2}{b}" but in a given question it was given as "\\frac{a^2}{b^2}" .. This is a mistake.








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